[proofplan]
We perturb the minimizer $u$ in the admissible direction $v-u$ by considering the segment $w_t = u+t(v-u)$ for $0 \leq t \leq 1$. Convexity of the admissible set keeps the whole segment inside $\mathcal K_{g,\psi}$, so the energy along this segment has a one-sided minimum at $t=0$. Expanding the energy as a quadratic polynomial in $t$ and using the one-sided minimality gives the desired variational inequality after dividing by $t>0$ and passing to the limit $t \downarrow 0$.
[/proofplan]
[step:Build admissible one-sided variations inside $\mathcal K_{g,\psi}$]
Fix $v \in \mathcal K_{g,\psi}$. For each $t \in [0,1]$, define the Sobolev function
\begin{align*}
w_t: U \to \mathbb{R}, \qquad w_t(x) := u(x) + t(v(x)-u(x)).
\end{align*}
Equivalently, $w_t = (1-t)u + tv$. Since $H^1(U)$ is a [vector space](/page/Vector%20Space), $w_t \in H^1(U)$. Moreover,
\begin{align*}
w_t - g = (1-t)(u-g) + t(v-g).
\end{align*}
Because $u-g \in H^1_0(U)$, $v-g \in H^1_0(U)$, and $H^1_0(U)$ is a vector subspace of $H^1(U)$, we have $w_t-g \in H^1_0(U)$.
It remains to check the obstacle constraint. Since $u \geq \psi$ and $v \geq \psi$ $\mathcal L^n$-a.e. on $U$, there is a Borel set $N \subset U$ with $\mathcal L^n(N)=0$ such that $u(x)\geq \psi(x)$ and $v(x)\geq \psi(x)$ for every $x \in U\setminus N$. For $x \in U\setminus N$ and $t \in [0,1]$,
\begin{align*}
w_t(x) = (1-t)u(x)+tv(x) \geq (1-t)\psi(x)+t\psi(x)=\psi(x).
\end{align*}
Thus $w_t \in \mathcal K_{g,\psi}$ for every $t \in [0,1]$.
[/step]
[step:Use minimality of $u$ along the admissible segment]
Define the one-variable energy profile
\begin{align*}
\Phi: [0,1] \to \mathbb{R}, \qquad \Phi(t) := E[w_t].
\end{align*}
Since $w_t \in \mathcal K_{g,\psi}$ for every $t \in [0,1]$ and $u=w_0$ minimizes $E$ over $\mathcal K_{g,\psi}$, we have
\begin{align*}
\Phi(t) \geq \Phi(0)
\end{align*}
for every $t \in [0,1]$.
[/step]
[step:Expand the energy as a quadratic polynomial in the variation parameter]
Let $h: U \to \mathbb{R}$ be the Sobolev function defined by $h := v-u$. Since $u,v \in H^1(U)$, we have $h \in H^1(U)$, and $\nabla h = \nabla v - \nabla u$ in $L^2(U;\mathbb{R}^n)$. For $t \in [0,1]$, the weak gradient is
\begin{align*}
\nabla w_t = \nabla u + t\nabla h.
\end{align*}
Therefore,
\begin{align*}
\Phi(t) = \frac{1}{2}\int_U |\nabla u + t\nabla h|^2\,d\mathcal L^n(x).
\end{align*}
Expanding the Euclidean square pointwise and integrating gives
\begin{align*}
\Phi(t) = \frac{1}{2}\int_U |\nabla u|^2\,d\mathcal L^n(x) + t\int_U \nabla u \cdot \nabla h\,d\mathcal L^n(x) + \frac{t^2}{2}\int_U |\nabla h|^2\,d\mathcal L^n(x).
\end{align*}
The three integrals are finite because $\nabla u,\nabla h \in L^2(U;\mathbb{R}^n)$ and the cross term is integrable by the [Cauchy-Schwarz inequality](/theorems/432) in $L^2(U;\mathbb{R}^n)$.
[guided]
The admissible path $w_t=u+t(v-u)$ converts the minimization problem over a convex set into a one-variable problem. Set $h:=v-u$, so $h \in H^1(U)$ and $\nabla h=\nabla v-\nabla u$ in $L^2(U;\mathbb{R}^n)$. For each $t \in [0,1]$,
\begin{align*}
\nabla w_t = \nabla u + t\nabla h.
\end{align*}
Now compute the energy along this path:
\begin{align*}
\Phi(t) = E[w_t] = \frac{1}{2}\int_U |\nabla u+t\nabla h|^2\,d\mathcal L^n(x).
\end{align*}
The pointwise Euclidean identity $|a+tb|^2=|a|^2+2t\,a\cdot b+t^2|b|^2$ applies with $a=\nabla u(x)$ and $b=\nabla h(x)$. Hence
\begin{align*}
\Phi(t) = \frac{1}{2}\int_U |\nabla u|^2\,d\mathcal L^n(x) + t\int_U \nabla u \cdot \nabla h\,d\mathcal L^n(x) + \frac{t^2}{2}\int_U |\nabla h|^2\,d\mathcal L^n(x).
\end{align*}
This expansion is legitimate because $\nabla u,\nabla h \in L^2(U;\mathbb{R}^n)$. The square terms are integrable by definition of $H^1(U)$, and the mixed term is integrable since the Cauchy-Schwarz inequality gives
\begin{align*}
\int_U |\nabla u \cdot \nabla h|\,d\mathcal L^n(x) \leq \|\nabla u\|_{L^2(U)}\|\nabla h\|_{L^2(U)} < \infty.
\end{align*}
Thus $\Phi$ is exactly a quadratic polynomial in the scalar parameter $t$, with linear coefficient equal to the variational expression we want to prove is nonnegative.
[/guided]
[/step]
[step:Extract the nonnegative first variation at $t=0$]
From $\Phi(t)\geq \Phi(0)$ for every $t \in [0,1]$ and the quadratic expansion above, for every $t \in (0,1]$ we obtain
\begin{align*}
t\int_U \nabla u \cdot \nabla h\,d\mathcal L^n(x) + \frac{t^2}{2}\int_U |\nabla h|^2\,d\mathcal L^n(x) \geq 0.
\end{align*}
Dividing by $t>0$ gives
\begin{align*}
\int_U \nabla u \cdot \nabla h\,d\mathcal L^n(x) + \frac{t}{2}\int_U |\nabla h|^2\,d\mathcal L^n(x) \geq 0.
\end{align*}
Since $\int_U |\nabla h|^2\,d\mathcal L^n(x)<\infty$, letting $t \downarrow 0$ yields
\begin{align*}
\int_U \nabla u \cdot \nabla h\,d\mathcal L^n(x) \geq 0.
\end{align*}
Substituting $h=v-u$ gives
\begin{align*}
\int_U \nabla u \cdot \nabla(v-u)\,d\mathcal L^n(x) \geq 0.
\end{align*}
Because $v \in \mathcal K_{g,\psi}$ was arbitrary, the variational inequality holds for every admissible competitor $v$.
[/step]
